获取向量&lt;Derived*&gt;转换为期望向量<

问题描述

考虑这些类.

class Base
{
   ...
};

class Derived : public Base
{
   ...
};

这个功能

void BaseFoo( std::vector<Base*>vec )
{
    ...
}

最后是我的向量

std::vector<Derived*>derived;

我想将 derived 传递给 BaseFoo 函数，但编译器不允许我这样做.如何解决这个问题，而不将整个向量复制到 std::vector?

I want to pass derived to function BaseFoo, but the compiler doesn't let me. How do I solve this, without copying the whole vector to a std::vector<Base*>?

推荐答案

vector 和 vector 是不相关的类型，所以你不能这样做.这在 C++ FAQ 这里 中有解释.

vector<Base*> and vector<Derived*> are unrelated types, so you can't do this. This is explained in the C++ FAQ here.

您需要将变量从 vector 更改为 vector 并插入 Derived 对象进入它.

You need to change your variable from a vector<Derived*> to a vector<Base*> and insert Derived objects into it.

此外，为了避免不必要地复制 vector，您应该通过常量引用而不是值来传递它:

Also, to avoid copying the vector unnecessarily, you should pass it by const-reference, not by value:

void BaseFoo( const std::vector<Base*>& vec )
{
    ...
}

最后，为了避免内存泄漏，并使您的代码异常安全，请考虑使用旨在处理堆分配对象的容器，例如:

Finally, to avoid memory leaks, and make your code exception-safe, consider using a container designed to handle heap-allocated objects, e.g:

#include <boost/ptr_container/ptr_vector.hpp>
boost::ptr_vector<Base> vec;

或者，更改向量以保存智能指针而不是使用原始指针:

Alternatively, change the vector to hold a smart pointer instead of using raw pointers:

#include <memory>
std::vector< std::shared_ptr<Base*> > vec;

或

#include <boost/shared_ptr.hpp>
std::vector< boost::shared_ptr<Base*> > vec;

在每种情况下，您都需要相应地修改您的 BaseFoo 函数.

In each case, you would need to modify your BaseFoo function accordingly.

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